Probability question involving bayesian theorem and Poisson's law (I think)

Question

"A yogurt factory has 10 cuves to produce yogurt. There are 2 cuves of type A, which produce yogurt that has, on average, 2.5 lumps per liter. There are 3 cuves of type B, producing on average 4 lumps per liter, and 5 cuves of type C, with on average 6 lumps per liter. In a pot (pot 1) containing a liter of yogurt without a label, there are 3 lumps. What is the probability that another pot (pot 2) coming from the same cuve contains 4 lumps"?

I'll just label "pot 1 contains 3 lumps" as p1=3, etc... If I use the Bayesian theorem I have:

$$P(p2=4 \mid p1=3) = \frac{P(p2=4) * P(p1=3 \mid p2=4)}{P(p2=4) * P(p1=3 \mid p2=4) + P(p2≠4) * P(p1=3 \mid p2≠4)}$$

I know how to calculate P(p2=4) and P(p2≠4) using Poisson's law, but obviously for the other terms I am not much more advanced... Of course developping them using Baye's theorem again seems pointless. What can I do?

Answer 1

The question is ill-posed. It can't be answered in this form, since we don't know how the cuve that the pots came from was chosen.

Assuming that the author of the question forgot to state that the cuve was chosen uniformly at random from the $10$ cuves described, the question can be answered as follows:

Denote by $C_i$ the event that the pot came from cuve $i$, by $n_i$ the number of cuves of type $i$ and by $\lambda_i$ the lump rate of cuves of type $i$. Then

\begin{eqnarray*} P(p_2=4\mid p_1=3) &=& \frac{P(p_2=4\land p_1=3)}{P(p_1=3)} \\ &=& \frac{\sum_in_iP(p_2=4\land p_1=3\mid C_i)}{\sum_in_iP(p_1=3\mid C_i)} \\ &=& \frac{\sum_in_iP(p_2=4\mid C_i)P(p_1=3\mid C_i)}{\sum_in_iP(p_1=3\mid C_i)} \\ &=& \frac{\frac1{3!4!}\sum_in_i\lambda_i^3\mathrm e^{-\lambda_i}\lambda_i^4\mathrm e^{-\lambda_i}}{\frac1{3!}\sum_in_i\lambda_i^3\mathrm e^{-\lambda_i}} \\ &=& \frac{\sum_in_i\lambda_i^7\mathrm e^{-2\lambda_i}}{4!\sum_in_i\lambda_i^3\mathrm e^{-\lambda_i}} \\ &=& \frac{2\cdot2.5^7\cdot\mathrm e^{-2\cdot2.5}+3\cdot4^7\cdot\mathrm e^{-2\cdot4}+5\cdot6^7\cdot\mathrm e^{-2\cdot6}}{4!\left(2\cdot2.5^3\cdot\mathrm e^{-2.5}+3\cdot4^3\cdot\mathrm e^{-4}+5\cdot6^3\cdot\mathrm e^{-6}\right)} \\ &\approx& 0.1585\;. \end{eqnarray*}

Answer 2

Your statement of Bayes's theorem seems to have an equation in the denominator of the right-hand side, which is certainly not a standard way of writing it. A more conventional expression would be the following: \begin{align} P(p_2&=4\left|p_1=3\right.)=\frac{P\left(p_2=4\right)P\left(p_1=3\left|p_2=4\right.\right)}{P\left(p_3=3\right)}\\ &= \frac{P\left(p_2=4\right)P\left(p_1=3\left|p_2=4\right.\right)}{P\left(p_2=4\right)P\left(p_1=3\left|p_2=4\right.\right)+ P\left(p_2\ne4\right)P\left(p_1=3\left|p_2\ne4\right.\right)}\ . \end{align} However, I don't see how this particular identity would be of much use for answering the question posed. For one thing, the conditional probability on the left-hand side doesn't include the information that the two pots come from the same cuve.

The conditional probability you should be calculating is \begin{align} &P\left(p_2=4\left|\left\{p_1=3\right\} \wedge \left\{\text{Cuve(pot2)}= \text{Cuve(pot1)}\right\}\right.\right)\\ &=\frac{P\left(\left\{p_2=4\right\}\wedge\left\{p_1=3\right\} \wedge \left\{\text{Cuve(pot2)}= \text{Cuve(pot1)}\right\}\right)}{P\left(\left\{p_1=3\right\} \wedge \left\{\text{Cuve(pot2)}= \text{Cuve(pot1)}\right\}\right)}\ . \end{align} You can't calculate this, however, without further information not given in the statement of the question. If you assume that the two pots are independently, and equally likely, to have come from each of the $10$ cuves, then the numerator can be written as \begin{align} &P\left(\left\{p_2=4\right\}\wedge\left\{p_1=3\right\} \wedge \left\{\text{Cuve(pot2)}= \text{Cuve(pot1)}\right\}\right)\\ &=\frac{1}{100}\sum_{i=1}^{10}P(p_2=4|\text{cuve}_i) P(p_1=3|\text{cuve}_i)\ , \end{align} and the denominator as \begin{align} &P\left(\left\{p_1=3\right\} \wedge \left\{\text{Cuve(pot2)}= \text{Cuve(pot1)}\right\}\right)\\ &=\frac{1}{100}\sum_{i=1}^{10}P(p_1=3|\text{cuve}_i)\ , \end{align} giving \begin{align} &P\left(p_2=4\left|\left\{p_1=3\right\} \wedge \left\{\text{Cuve(pot2)}= \text{Cuve(pot1)}\right\}\right.\right)\\ &=\frac{\sum_{i=1}^{10}P(p_2=4|\text{cuve}_i) P(p_1=3|\text{cuve}_i)}{\sum_{i=1}^{10}P(p_1=3|\text{cuve}_i)}\ . \end{align}